Question

有数据框df有两列，datetime.time对象如下：

  TimeA           TimeB
  00:50:13        00:50:00
  00:51:46        00:50:00
  00:52:58        00:50:00
  00:54:05        00:51:00

我想创建第三列，这两列之间存在差异。列中的元素是datetime.time对象。首先，我尝试了以下方法中的单个值：

>from datetime import datetime, date, time
>TimeA = datetime.combine(datetime.min, df.iloc[0,0]) - datetime.min
>TimeB = datetime.combine(datetime.min, df.iloc[0,1]) - datetime.min
> diff = TimeA - TimeB

它给出了以下结果：

datetime.timedelta(0, 13)

但是，当我尝试转换整列时：

df['TimeA_2'] = df['TimeA'].apply(lambda x : datetime.combine(date.min, x) - datetime.min)

发生以下错误：

combine() argument 2 must be datetime.time, not float

没有意义，因为当我检查两列中的元素类型时，它们是datetime.time。我不知道，错误在哪里。非常感谢任何帮助。

Answer 1

您可以将列转换为datetime个对象，然后获取差异并转换为分钟：

import numpy as np

df[['TimeA', 'TimeB']] = df[['TimeA', 'TimeB']].apply(pd.to_datetime)

#                 TimeA               TimeB
# 0 2018-03-05 00:50:13 2018-03-05 00:50:00
# 1 2018-03-05 00:51:46 2018-03-05 00:50:00
# 2 2018-03-05 00:52:58 2018-03-05 00:50:00
# 3 2018-03-05 00:54:05 2018-03-05 00:51:00

df['Diff'] = (df['TimeA'] - df['TimeB']) / np.timedelta64(1, 'm')

#                 TimeA               TimeB      Diff
# 0 2018-03-05 00:50:13 2018-03-05 00:50:00  0.216667
# 1 2018-03-05 00:51:46 2018-03-05 00:50:00  1.766667
# 2 2018-03-05 00:52:58 2018-03-05 00:50:00  2.966667
# 3 2018-03-05 00:54:05 2018-03-05 00:51:00  3.083333

Answer 2

IIUC使用pd.to_timedelta：

df[['TimeA','TimeB']] = df[['TimeA','TimeB']].apply(pd.to_timedelta)
df['Diff'] = (df['TimeA'] - df['TimeB'])

输出：

     TimeA    TimeB     Diff
0 00:50:13 00:50:00 00:00:13
1 00:51:46 00:50:00 00:01:46
2 00:52:58 00:50:00 00:02:58
3 00:54:05 00:51:00 00:03:05

OR

df['Diff'] = (df['TimeA'] - df['TimeB']).dt.total_seconds() / 60

输出：

     TimeA    TimeB      Diff
0 00:50:13 00:50:00  0.216667
1 00:51:46 00:50:00  1.766667
2 00:52:58 00:50:00  2.966667
3 00:54:05 00:51:00  3.083333

在数据框中减去两个datetime.time列

2 个答案: